[Resolved] Conditional Block on content template does not render the Single Field.

[Author]

Posts

[darshanP]

I am trying to: display a single field inside a conditional block on a content template.

Link to a page where the issue can be seen: hidden link

I expected to see: The value stored in that custom field.

Instead, I got: An unrendered short code on the front end.

[Luo Yang]

Hello,

I assume we are talking about the shortcode [types field='category' output='normal' ][/types] in the URL you mentioned above.

I have tried it in my localhost, there isn't the problem you mentioned above.

Since it is a shortcode, please put it into a shortcode block, and test again.

Also check these:
1) In case it is a compatibility problem, please deactivate all other plugins, and switch to wordpress default theme 2020, deactivate all custom PHP/JS code snippets, and test again

2) If the problem still persists, please provide database dump file(ZIP file) of your website, thanks
https://toolset.com/faq/provide-supporters-copy-site/

[darshanP]

Any update?

[Luo Yang]

I am downloading the files, will update here if there is anything found.

[Luo Yang]

Here are what I found, please edit the content template "single-incident-entry":
hidden link

in the problem row(just below text "Attachment"), you should be able to see message: this block contains unexpected or invalid content.

It conducts the problem.

I suggest you try one of below options:
1) Convert it to HTML block, and follow our document to setup the [wpv-conditional] shortcode
hidden link

For example:

[wpv-conditional if="  ( ( '[wpv-post-field name="entry_type" ]' = 'Request' ) ) " ]
...
[/wpv-conditional]

2) Follow our document to setup the conditional block, and test again.
https://toolset.com/documentation/user-guides/gutenberg/using-toolset-conditional-block/

[darshanP]

I was able to figure it out with your help.

Using the Toolset Layout Builder the conditional field does not work, you have to type it out manually using the short code.